AQA MFP4 w TSM EX JUN08

Note, next, that the working in ii is correct on a follow-through basis and both marks are scored as aresult.. It is then strange to see that the next vector product attempted by this ca

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Teacher Support Materials

2008 Maths GCE

Paper Reference MFP4

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12 7

equation (actually, two equations, but they are in fact the same line) satisfied by the components ofany representative eigenvector, and then any suitable eigenvector has been written down [Note thatany (non-zero) multiple of the given answers would also have been a valid answer.]

Mark scheme

Question 2

The vectors a, b and c are given by

a = i + 2j + 3k , b = 2i + j + 2k and c = – 2i + tj + 6k

where t is a scalar constant.

(a) Determine, in terms of t where appropriate:

(b) Find the value of t for which a, b and c are linearly dependent. (2 marks)

(c) Find the value of t for which c is parallel to a  b (2 marks)

Student response

This is a very interesting script, and there are several worthwhile points to be drawn from it In (a) (i),

this candidate has attempted the vector product using the Distributive Law – as one would if

multiplying out two brackets – then applying the properties that i  i = 0, i  j = k and i  k = – j

(etc.) However, by employing just about the longest possible method, the whole thing has gone badlywrong, and none of the components in the answer vector is correct

Note, next, that the working in (ii) is correct on a follow-through basis and both marks are scored as aresult It is then strange to see that the next vector product attempted by this candidate, in (iii), uses amuch more concise method than he employed in (i); however, by this point, follow-through of possiblyone or two previous follow-throughs has not been considered suitable in the mark-scheme and,although correct on this basis, only the method mark has been allowed In (b), however, full follow-through of (a) (ii)’s answer has been permitted

Finally, there are a couple of issues that arise in part (c) The first is that (a) (iii)’s answer hasmysteriously changed In cases of this sort of carelessness, follow-through rules cannot apply, and thiscandidate would necessarily find himself penalised for inconsistent working and/or answers The much

greater point at stake here, however, is that the candidate HAS actually recognised what should be going on … BUT has then failed to realise that this clearly shows that an error has occurred earlier

and, rather than making an obviously incorrect statement, he should have gone back and checked hisprevious working in order to put it right

Mark Scheme

3 1 1

1 1 1

, where k is a constant.

Determine, in terms of k where appropriate,

Student Response

[Note: this candidate’s (a) was fully correct and has not been included here.]

The marks for this part of the question are split between the various types There are two method (M)

marks; the first for the general idea of attempting the use of the “transposed matrix of co-factors”approach – which this candidate has clearly attempted – and the second for the use of the alternating

signs within the matrix of co-factors and the transposition This second method mark has not been

earned as this candidate has reflected the matrix’s elements about the non-leading

(top-right-to-bottom-left) diagonal, instead of the leading diagonal (top-left-to-bottom-right) If you look at the

mark-scheme, you will see that the first of the two accuracy (A) marks is awarded for at least five correct

entries Thus, forgetting to change the signs in alternate places can still lead to the acquisition of this

first A mark However, messing up the transposition throws both of these marks away.

The final mark is a B mark, which is a stand-alone mark for correct application of a single result, idea

or method, often one that can be done at any stage of a solution.Here, the transposed matrix of factors has to be multiplied by the reciprocal of the determinant, which was the answer to part (a) Inthis solution, this is the factor of ½ that appears on the very last line Even had the candidate got thiswrong in (a), this would have followed-through here (except for the obviously problematic answer ofzero)

co-Mark Scheme

= 7.

(a) Find, to the nearest 0.1o, the acute angle between the two planes (4 marks)

(b) (i) The point P(0, a, b) lies in both planes Find the value of a and the value of b.

Student Response

Although there is very little explanation of what is going on here, this candidate’s answers are clearlyset out and obviously attempting to do the right things In (a), however, you can see that a lack of initialexplanation has led to their getting carried away; having actually found the correct angle, they then

proceed to find another one You may have noticed the marking principle of ISW – “ignore subsequent

working” – which is often applied to candidates’ working when, having gained an acceptable correct

form of an answer, they then proceed to “tidy it up” incorrectly ISW applies in such cases It, sadly,

doesn’t apply when the working clearly proceeds in an incorrect way; in this case, to the finding of

another angle altogether

The rest of the question runs very smoothly, although this candidate has clearly failed to notice thatthe required answers to (b) (i) and (ii) are actually the bits of the answer needed for (iii), and hasstarted all over again using a different method As it is correct, full marks have been gained, but someprecious time may have been used up in this way

Mark Scheme

Question 5

A plane transformation is represented by the 2  2 matrix M The eigenvalues of M are

1 and 2, with corresponding eigenvectors 

1

respectively.

(a) State the equations of the invariant lines of the transformation and explain which of these

(b) The diagonalised form of M is M = U D U– 1, where D is a diagonal matrix.

(i) Write down a suitable matrix D and the corresponding matrix U. (2 marks)

(iii) Show that Mn= 

1 ) ( 1

n

n

for all positive integers n, where f(n) is a

Student Response

.

This candidate has made a real mess of part (a), where the invariant lines of the transformation were

to be deduced from the information given about eigenvalues and eigenvectors Also, it is really no useguessing which of these invariant lines is a line of invariant points, as no marks are given for luckyguessers, and so it is essential to give a valid reason in these cases He then proceeds correctly intopart (b), but falls down at the point when he has to write down the inverse of a 22 matrix This isstrange, because this candidate had previously gained full marks on Q3, where the question asked forthe inverse of a much more difficult, 33 matrix

This silly mistake has also cost him an accuracy mark in the final part Moreover, his final answer

clearly doesn’t match his answer for M1 a few lines above Surely this should flag up that a bit of

careful checking is needed somewhere; at the least a check that his U and U– 1 multiply to give the

identity matrix, I, would have been worth the time and effort

Mark Scheme

Question 6

Three planes have equations

4 2

5

3 4

x

z y

x

b z

y x

where a and b are constants.

(a) Find the coordinates of the single point of intersection of these three planes in the

(b) (i) Find the value of a for which the three planes do not meet at a single point.

(3 marks)

(ii) For this value of a, determine the value of b for which the three planes share a

Student Response

I have included this candidate’s solution to this question as it is absolutely fantastic Every step hasbeen explained, and every bit of working is clearly laid out Now, if only all candidates presented theirworking like this!

Mark Scheme

2 0 2

1 1 3

.

(a) (i) Evaluate det W, and describe the geometrical significance of the answer in relation

(b) The plane  has equation r

1

= 0.

(ii) The point P has coordinates (a , b , c) Show that, whatever the values of a, b and c,

Student Response

Despite scoring the majority of the marks on this question, this candidate has made small, but costly,mistakes which could have been discovered with either some more careful thought, or by a bit ofchecking His first answer in (a) is correct, but he is then unsure how to interpret it At first, he refers to

“area”; then elects to hedge his bets by crossing it out (it should refer to volume of course) Having got

in a muddle about it, he then suggests that the scale factor (of something unspecified) is 1, whichdoesn’t even match his answer of 0

To find the (cubic) characteristic equation for the given matrix, one could elect to expand this (correct)determinant fully and then work with the resulting expression algebraically, or to undertake somerow/column operations first in order to simplify it This candidate chooses the first approach, but then

“spots” a common factor of (3 – ) which doesn’t even appear in the final term of his expansion.Factorising the following incorrect quadratic term, or solving the equation by (say) the quadraticformula, is now as much a matter of luck as anything, and this candidate has earned only the two

method marks available, but none of the accuracy marks This despite the fact that the zero

determinant in (a) should have flagged up the fact that (at least) one eigenvalue is zero

The final part, though not explained, is fully correct

Mark Scheme

Question 8

By considering the determinant

x z y

y x z

z y x

, show that (x + y + z) is a factor of x3+ y3+ z3– kxyz

Student Response

Commentary

There are essentially two parts to this question Firstly, expand the determinant fully as it stands to get

the expression x3 + y3 + z3 – 3xyz Then use row/column operations to extract the rather obvious factor of (x + y + z) This candidate has done both of these things, and yet not gained the final mark.

This final mark is for putting the two halves together, and stating what must seem the obvious that,

since A is a factor of ∆(the determinant), and ∆= B , then A is a factor of B Alternatively, any line

of working that began with AC = … and ended up with … = B would also have done the trick.

However, since the question actually gave you this up front, you need to be careful to ensure that yourworking dots all the i’s and crosses all the t’s in the right way Leaving the two sides of a line ofreasoning un-matched will usually lose you a mark Especially on a further maths paper

Mark Scheme